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The Relationship Between Group
Representations and Their Generators

An abtract group can be represented by square matrices and their multiplication.

Consider square matrices with complex elements. Let U be a matrix such that

U^{†}U = I

where U^{†} is the transpose of the complex conjugate of U and I is the appropriate identity matrix. A matrix satisfying the above
condition is said to be unitary. Let U(α) be a continuum of matrices indexed by the parameter α. Let U(0)=I.

A matrix G is said to be an infinitesimal generator of the U(α) if

U(dα) = I + iGdα

where i is the imaginary unit √-1 and dα is an infinitesimal.

Cosider now what the transpose of the complex conjugate of U(dα) would be. The conjugate of the sum of two matrices is the sum of the conjugates. Likewise the transpose of the sum of matrices is the sum of the transposes.
The identity matrix is real and symmetric so it is equal to its transpose conjugate. The conjugate of an expression is that expression with i replaced by
−i. Thus

U(dα)^{†} = I − iG^{†}dα

Now consider

U(dα)^{†}U(dα) = (I − iG^{†}dα)(I + iGdα) =
I − iG^{†}dα + iGdα
since dα is an infinitesimal

Then

U(dα)^{†}U(dα) = I
means that
I − iG^{†}dα + iGdα = I
and hence
i(G−G^{†})dα = 0 which reduces to
G^{†} = G

A matrix satisfying this condition is said to be Hermitian.

Special Unitary Matrices

A matrix which is unitary and also has a determinant equal to 1 is said to be a special unitary matrix.

There is a theorem for determinants that says

det(exp(A)) = exp(tr(A)

Let

U = exp(iG)
and
det(U) = 1

Then by the above theorem concerning determinants

det(U) = det(exp(iG)) = exp(tr(iG)) = exp(i·tr(G)) = 1
which means that
i·tr(G) = 0
and hence
tr(G) = 0

Thus the generators for the elements of a special unitary group are Hermitian and have traces of zero.